Question

A closed, rigid tank contains a two‐phase liquid–vapor mixture of Refrigerant 22 initially at −20°C with a quality of 50.36%. Energy transfer by heat into the tank occurs until the refrigerant is at a final pressure of 6 bar. Determine the final temperature, in °C. If the final state is in the superheated vapor region, at what temperature, in °C, does the tank contain only saturated vapor?

Answer 1

Step-by-step explanation:

The given data is as follows.

Initial temperature of the system (
`T_(1)`) =
`-20^(o)C`

So, at state 1 quality of the system is (
`x_(t)`) = 0.5036

Pressure at state 2, (
`P_(2)`) = 6 bar

As, it is a rigid tank hence, the specific volumes at state 1 and state 2 are equal.

So,
`\\u_(2) = \\u_(t)`

Now, from the saturated refrigerant-22 table taking specific volume at
`-22^(o)C` is as follows.

`(\\u_(f))_(t) = 0.7427 * 10^(-3) m^(3)/kg`

`(\\u_(g))_(t) = 0.0926 m^(3)/kg`

Thus,
`\\u_(2) = \\u_(t)`

=
`(\\u_(f))_(t) + x_(t) ((\\u_(g))_(t) - (\\u_(f))_(t))`

= 0.0007427 + 0.5036 (0.0926 - 0.0007427)

=
`0.04700 m^(3)/kg`

Hence,
`\\u_(g)` =
`0.04700 m^(3)/kg`

Super heated refrigerant-22 tables take pressure at 6 bar. Interpolation method used to find the temperature is as follows.

`\\u_(t) = 0.04628 m^(3)/kg` and
`T_(t) = 40^(o)C`

`\\u_(t) = 0.04724 m^(3)/kg` and
`T_(2) = 45^(o)C`

`T_(vap) = T_(t) + ((\\u_(g) - \\u_(t))(T_(2) - T)_(t))/(\\u_(2) - \\u_(t))`

=
`40^(o)C + ((0.04700 - 0.04727) * (45 - 40))/(0.04724 - 0.04628)`

= 40 + 3.75

=
`43.75^(o)C`

Saturated vapor refrigerant-22 takes specific volume at
`0.04700 m^(3)/kg`.

Therefore, interpolation method used to find the temperature will be as follows.

`\\u_(t) = 0.0492 m^(3)/kg` and
`T_(t) = -1.45^(o)C`

`\\u_(2) = 0.0469 m^(3)/kg` and
`T_(2) = 0.12^(o)C`

`T_{\text{saturated vapor}} = T_(t) + ((\\u_(g) - \\u_(t))(T_(2) - T_(t))/(\\u_(2) - \\u_(t))`

=
`-1.45^(o)C + ((0.04700 - 0.0492) * (0.12 - (-1.45)))/(0.0469 - 0.0492)`

= -1.45 + 1.50

=
`0.051^(o)C`

Thus, we can conclude that at
`0.051^(o)C` tank contains only saturated vapor.

Answer 2

To find the final temperature, we use the saturation table for Refrigerant 22. The final state is in the superheated vapor region and the final temperature is 39.3°C when the tank contains only saturated vapor.

To find the final temperature of the two-phase liquid-vapor mixture of Refrigerant 22 in the closed, rigid tank, we need to use the saturation table for Refrigerant 22. From the given initial temperature of -20°C and initial quality of 50.36%, we can determine that the initial state is in the two-phase region. Using the saturation table, we find that the saturation temperature at the given initial pressure is -10.9°C. Since the final pressure is given as 6 bar, which is above the critical pressure of Refrigerant 22, the final state will be in the superheated vapor region.

To determine the final temperature, we need to find the corresponding enthalpy value for dry saturated vapor at 6 bar pressure. Consulting the saturation table again, we find that the enthalpy of dry saturated vapor at 6 bar is 275.92 kJ/kg. Therefore, the final temperature of the tank containing saturated vapor is 39.3°C.

In conclusion, the final temperature of the two-phase liquid-vapor mixture is determined to be in the superheated vapor region and is found to be 39.3°C when the tank contains only saturated vapor.